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Model Answer B

The document contains calculations to solve multiple physics problems involving forces and moments. It finds the magnitude and direction of unknown forces. It also calculates the torque and resultant moment of several couples and forces.

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Ahmed Hamed
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33 views2 pages

Model Answer B

The document contains calculations to solve multiple physics problems involving forces and moments. It finds the magnitude and direction of unknown forces. It also calculates the torque and resultant moment of several couples and forces.

Uploaded by

Ahmed Hamed
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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Model Answer (B)

Rx = Fcos + 600 cos 45+ 325 (5/13) = 800 (1)

Ry = Fsin + 325 (12/13) - 600 sin 45 = 0 (2)

From (1), (2) F = 279.8 N ,  = 26.4

F300N = 300 cos 30 sin 45 i + 300 cos 30 cos 45 j – 300 sin 30 k = 183.7i+183.7j-150k

F180N = -180i

FR = 400j

Funknown force = – 3.7i + 216.3j + 150k with magnitude and direction

−3.7 2 + 216.3 2 + 150 2 =263.2 N


−3.7 216.3 150
∝= 𝑐𝑜𝑠 −1 = 90.8° , 𝛽 = 𝑐𝑜𝑠 −1 = 34.7°, 𝛾 = 𝑐𝑜𝑠 −1 = 55.3°
263.2 263.2 263.2

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3𝑖+2𝑗 −6𝑘
F = 70 ∗ = 30𝑖 + 20𝑗 − 60𝑘
7

FED = 30𝑖 + 20𝑗 − 60𝑘 ∙ 0.6𝑖 − 0.8𝑗 = 2 𝑁

F ED= 70 2 − 2 2 = 69.9 N

FED = 1.2i-1.6j

F ED = 28.8i+21.6j-60k

MO = 𝑂𝐴 × 𝐹 = 6𝑗 + 6𝑘 × 30𝑖 + 20𝑗 − 60𝑘 = – 480 i + 180 j – 180 k

MD = 𝐷𝐴 × 𝐹 = −3𝑖 + 4𝑗 + 6𝑘 × 30𝑖 + 20𝑗 − 60𝑘 = – 360 i – 180 k

MED = 𝑀𝐷 ∙ 𝑢𝐸𝐷 𝑢𝐸𝐷 = −360𝑖 − 180𝑘 ∙ 0.6𝑖 − 0.8𝑗 0.6𝑖 − 0.8𝑗 = – 129.6 i + 172.8 j

MCD = 0

MOC = 𝑀𝑜 ∙ 𝑢𝑂𝐶 𝑢𝑂𝐶 = −480𝑖 + 180𝑗 − 180𝑘 ∙ 𝑘 𝑘 = −180𝑘

MCB = 0
Couples:

Let 10 N = 5N  + 5N  and 35N = 15N  + 20N

+5*8=+40, -5*12=-60Nm, +15*5=+75Nm, -20*8=-160Nm, +10*13=+130Nm,

Total = +25Nm

MO = MA = MB = ME = +25 Nm

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M100N Couple = 2𝑖 − 3.464𝑗 + 5𝑘 × 100𝑗 = −500𝑖 + 200𝑘

M50N Couple = 2𝑖 − 3.464𝑗 × 50𝑘 = −173.2𝑖 − 100𝑗

Mresultant Couple = −673.2𝑖 − 100𝑗 + 200𝑘

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